The Stacks project

Lemma 9.22.3. Let $L/K$ be a Galois extension with Galois group $G$. Let $\Lambda $ be the set of finite Galois subextensions, i.e., $\lambda \in \Lambda $ corresponds to $L/L_\lambda /K$ with $L_\lambda /K$ finite Galois with Galois group $G_\lambda $. Define a partial ordering on $\Lambda $ by the rule $\lambda \geq \lambda '$ if and only if $L_\lambda \supset L_{\lambda '}$. Then

  1. $\Lambda $ is a directed partially ordered set,

  2. $L_\lambda $ is a system of $K$-extensions over $\Lambda $ and $L = \mathop{\mathrm{colim}}\nolimits L_\lambda $,

  3. $G_\lambda $ is an inverse system of finite groups over $\Lambda $, the transition maps are surjective, and

    \[ G = \mathop{\mathrm{lim}}\nolimits _{\lambda \in \Lambda } G_\lambda \]

    as a profinite group, and

  4. each of the projections $G \to G_\lambda $ is continuous and surjective.

Proof. Every subfield of $L$ containing $K$ is separable over $K$ (follows immediately from the definition). Let $S \subset L$ be a finite subset. Then $K(S)/K$ is finite and there exists a tower $L/E/K(S)/K$ such that $E/K$ is finite Galois, see Lemma 9.16.5. Hence $E = L_\lambda $ for some $\lambda \in \Lambda $. This certainly implies the set $\Lambda $ is not empty. Also, given $\lambda _1, \lambda _2 \in \Lambda $ we can write $L_{\lambda _ i} = K(S_ i)$ for finite sets $S_1, S_2 \subset L$ (Lemma 9.7.5). Then there exists a $\lambda \in \Lambda $ such that $K(S_1 \cup S_2) \subset L_\lambda $. Hence $\lambda \geq \lambda _1, \lambda _2$ and $\Lambda $ is directed (Categories, Definition 4.21.4). Finally, since every element in $L$ is contained in $L_\lambda $ for some $\lambda \in \Lambda $, it follows from the description of filtered colimits in Categories, Section 4.19 that $\mathop{\mathrm{colim}}\nolimits L_\lambda = L$.

If $\lambda \geq \lambda '$ in $\Lambda $, then we obtain a canonical surjective map $G_\lambda \to G_{\lambda '}$, $\sigma \mapsto \sigma |_{L_{\lambda '}}$ by Lemma 9.21.8. Thus we get an inverse system of finite groups with surjective transition maps.

Recall that $G = \text{Aut}(L/K)$. By Lemma 9.22.2 the restriction $\sigma |_{L_\lambda }$ of a $\sigma \in G$ to $L_\lambda $ is an element of $G_\lambda $. Moreover, this procedure gives a continuous surjection $G \to G_\lambda $. Since the transition mappings in the inverse system of $G_\lambda $ are given by restriction also, it is clear that we obtain a canonical continuous map

\[ G \longrightarrow \mathop{\mathrm{lim}}\nolimits _{\lambda \in \Lambda } G_\lambda \]

Continuity by definition of limits in the category of topological groups; recall that these limits commute with the forgetful functor to the categories of sets and topological spaces by Topology, Lemma 5.30.3. On the other hand, since $L = \mathop{\mathrm{colim}}\nolimits L_\lambda $ it is clear that any element of the inverse limit (viewed as a set) defines an automorphism of $L$. Thus the map is bijective. Since the topology on both sides is profinite, and since a bijective continuous map of profinite spaces is a homeomorphism (Topology, Lemma 5.17.8), the proof is complete. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BU2. Beware of the difference between the letter 'O' and the digit '0'.